Problem: Divide the following complex numbers. $ \dfrac{40+10i}{-5+3i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-5-3i}$ $ \dfrac{40+10i}{-5+3i} = \dfrac{40+10i}{-5+3i} \cdot \dfrac{{-5-3i}}{{-5-3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(40+10i) \cdot (-5-3i)} {(-5+3i) \cdot (-5-3i)} = \dfrac{(40+10i) \cdot (-5-3i)} {(-5)^2 - (3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(40+10i) \cdot (-5-3i)} {(-5)^2 - (3i)^2} = $ $ \dfrac{(40+10i) \cdot (-5-3i)} {25 + 9} = $ $ \dfrac{(40+10i) \cdot (-5-3i)} {34} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({40+10i}) \cdot ({-5-3i})} {34} = $ $ \dfrac{{40} \cdot {(-5)} + {10} \cdot {(-5) i} + {40} \cdot {-3 i} + {10} \cdot {-3 i^2}} {34} $ Evaluate each product of two numbers. $ \dfrac{-200 - 50i - 120i - 30 i^2} {34} $ Finally, simplify the fraction. $ \dfrac{-200 - 50i - 120i + 30} {34} = \dfrac{-170 - 170i} {34} = -5-5i $